Dear Vipul,
Solution:- Given, AC = a BC and BC = a AB. We want to find AC: AB  = a2
Since, AC is the diameter of the circle, it subtends 900 angle at B. Let, AC = 'd' and angle C be 'x'. Then, 
AB = d sinx , and BC = d cosx
So, AC = a BC will give 1 = a cosx     or    cosx = 1/a

And, BC = a AB will give cos x = a sinx 
Since, cos x = 1/a, sin x = √(a2 -1) / a
So, 1 = a √(a2 -1)          or         a4 - a2 -1 = 0
On solving quad. eq., we'll get - a2 = (√5 +1) / 2  [ANS]

Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.

All the best Vipul!!!

Regards,
Askiitians Experts
Priyansh Bajaj


						

							
ΔABC is an right angle triangle as Ac is the diameter of the circle. 
Now it is given that AC-BC=BC-AB
AC+AB=2BC
AC=2BC-AB............(1)
AB²+BC²=AC². .........(Pythagoras theorem) 
AB²+BC²=(2BC-AB)²............(using eqn.  1)
AB²+BC²=4BC²+AB²-4*BC*AB
BC²-4BC²=-4*BC*AB
3BC²=4*BC*AB
3BC²/4BC=AB
3/4BC=AB........(2)
AB²+BC²=AC²
(3/4BC)²+BC²=AC².......(Using eqn. 2)
9/16BC²+BC²=AC²
9/16BC²+16/16BC²=AC²
25/16BC²=AC²
5/4BC=AC........(3)
AC:AB=AC/AB
(5/4BC)/(3/4BC)..........(Using eqn.  2 &3)
5/3=AC/AB=AC:AB